\begin{answer}
$$
    \begin{aligned}
\theta &= -\arg \min_\theta \log p(y|x, \theta)p(\theta)\\
&= \arg \min_\theta -\log p(y|x, \theta) - \log p(\theta)\\
&= \arg \min_\theta -\log p(y|x, \theta) - \log \frac{1}{2b}\exp(-\frac{\|\theta\|}{b})\\
&= \arg \min_\theta -\log p(y|x, \theta) + \frac{1}{b}\|\theta\|
\end{aligned}
$$

So here, $\lambda = 1/ b$.

\end{answer}
